All categories

3 years ago

Make sure your response is 3-5 sentences.

Physics

1 answer:

g100num [7]3 years ago

8 0

Answer: please see below

Explanation:

A manned space exploration is defined as the exploration of individuals --- astronauts in space using a spacecraft as a vehicle and are responsible for operating its controls

The extreme conditions in space that challenge manned space exploration is as follows.

1. extreme loud sound waves cause by the launch of spacecraft which can shatter the spacecraft

2. extreme Temperatures in space ranging from extreme hot temperatures (near the sun) to extreme cold temperatures ( below freezing point out of space.

3.micrometeorite showers responsible for sandblasting can damage spacecraft.

4.Ultra violet Radiation which can alter the control unit of the spacecraft

Because of theses extreme conditions that pose challenges to space explorations, necessary precautions should be taken into consideration to be able to overcome such challenges. These precautions include building the spacecraft and the control unit in such a way that can resist these harmful conditions, also taking in mind safe escape routes for the astronauts in case of failures.

You might be interested in

Two stones, one with twice the mass of the other, are thrown straight up and rise to the same height h. Compare their changes in

Lady bird [3.3K]

<h2>The gravitational potential energy is double for stone with twice the mass of other stone.</h2>

Explanation:

Let mass of stone 1 be m.

Mass of stone 2 is twice the mass of stone 1.

Mass of stone 2 = 2m

We know that

Gravitational potential energy = Mass x acceleration due to gravity x Height

PE = mgh

For stone 1 ,

PE₁ = mgh

For stone 2 ,

PE₂ = 2mgh = 2 PE₁

So the gravitational potential energy is double for stone with twice the mass of other stone.

4 0

3 years ago

Read 2 more answers

As Clinton walks he pushes his shoe against the track.

Valentin [98]

Answer:A:The track pushes back on Clinton's shoe with the same force.

Explanation:According to Newton's third law of motion, for every action force there is an equal and opposite reaction force. In this case, the action force is Clinton's shoe pushing on the track. As this happens, there is an equal and opposite reaction force in which the track pushes back on Clinton's shoe with the same force.

4 0

3 years ago

A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because of the cramped space, the force must be exerted upward

riadik2000 [5.3K]

Answer:

Torque, $\tau=34.6\ N.m$

Explanation:

It is given that,

Length of the wrench, l = 0.5 m

Force acting on the wrench, F = 80 N

The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :

$\tau=Fr\ sin\theta$

$\tau=80\times 0.5\ sin(60)$

$\tau=34.6\ N.m$

So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.

7 0

3 years ago

Describe how different types of motion are represented by distance-time graphs and velocity-time graphs.

eduard

Answer:

non-accelerated movement

velocity versus time a horizontal straight line.

distance versus time gives a horizontal straight line.

accelerated motion

graph of velocity versus time s an inclined line and the slope

graph of distance versus time is a parabola of the form

Explanation:

In kinematics there are two types of steely and non-accelerated movements

In a the velocity of the body is constant therefore a speed hook against time gives a horizontal straight line.

A graph of distance versus time is a straight line whose slope is the velocity of the body

x = v t

In an accelerated motion the velocity changes linearly with time, so a graph of velocity versus time is an inclined line and the slope is the value of the acceleration of the body

v = v₀ + a t

A graph of distance versus time is a parabola of the form

x =v₀ t + ½ a t²

4 0

3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

<u>For the course of upward acceleration:</u>

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

3 years ago